Prove that $V_j$ is finite-dimensional for each $j = 1,...m$ Suppose $V_1,\ldots,V_m$ are vector spaces that $V_1 \times \cdots \times V_m$ is finite dimensional.
Prove that $V_j$ is finite-dimensional for each $j = 1,\ldots, m$
So far I have considered proof by contradiction.
Suppose $V_j$ is infinite dimensional
 A: Hint: Let $p_1:V_1\times ...V_n\rightarrow V_1$ be the canonical projection.
If $e_1,...,e_p$ generates $V_1\times...\times V_m$, $p_1(e_1),...,p_1(e_p)$ generates $V_1$.
A: I can see two ways of doing this: One is the one I've hinted at in the comments section above, and it continues your line of thought for a proof by contradiction. However, I think that this is easier:

Given some $i \leq n$, the vector space $V_i$ is in a natural way a subspace of the product: for any vector $v\in V_i$, consider the element $(0,0,\ldots,0,v,0,\ldots,0)$ in the product.
Now, say the dimension of the product is $n$. Take any $n+1$ vectors $v_1, \ldots, v_{n+1} \in V_i$ and look at the corresponding vectors in the product. Since the product has dimension $n$ and there are $n+1$ vectors, they must be linearly dependent.
This shows that any $n+1$ vectors in $V_i$ are linearly dependent, and therefore $V_i$ has dimension at most $n$.
A: For a fixed $V_j:1\leq j\leq n\ $, choose a basis $\left \{ v^{(j)}_{i} \right \}_{1\leq i\leq n_{j}}$ and define $T:V_j\rightarrow V_1\times \cdots \times V_n$ by $v^{(j)}_{i}\mapsto (0,\cdots, v^{(j)}_{i},0\cdots,0) $. 
Now use the fact that $T$ is injective, and Im $T$ is a subspace of $V_1\times \cdots \times V_n$ to conclude that $V_j$ is finite dimensional. 
